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Comment Re:I'm no where near as smart as most of you.. (Score 1) 711

Look, there is no sense in continuing this debate. You're just not getting it, and then get all defensive and call us 'arrogant'. No, it's just that you refuse or are unable to understand the issue. Not seeing the 'entire picture', indeed.

Your analogies; turbines, flying wheels, etc. just prove that. You simply do not understand that those examples can not do the same as an EMdrive, because their power ratio is too weak; they never get to the point where you get MORE power out than you put in, because even for a photon rocket it would need to reach the speed of light for that, which is impossible. But since the reported thrust to power ratios for the EmDrive are orders of magnitude higher, you get far *more* energy back than you put in (at speeds that ARE attainable) - which is of course, bullocks. Losses, friction, torque, etc. only matter if you have the same amount of energy that you 'recycle'; that's why, in practise, you can not build a perpetuum mobile with the same energy you put in, because each time you get losses to the original amount of energy, and eventually it dries up. But those systems *can not* make more energy on itself than you put in, because their power ratio is too weak. No other device, not even a photon rocket, can achieve it with the power ratio it has. The EMdrive claims it has and thus does.

Ergo, you can get far more energy out of it than you put in, at speeds/thrusts that are attainable, and thus, it does not matter if it's from an outside source or not, as long as the energy keeps flowing. And it keeps flowing because it generates MORE energy than you put in, in total. With power ratios like that, you do not need to accelerate 'indefinitely', that is just the point!

Now, you can debate semantics all you want, claiming a perpetuum mobile means something else to you, but it doesn't change anything to the fact that such a thing isn't possible, and since the EMdrive would make it possible, the EMdrive, as a reactionless drive with the proclaimed power ratio, is nonsense. Yes, yes, you do not agree, but that still doesn't change anything to that fact.

Several people have tried to explain it to you in the most simple and clear way by now, and yet you persist in refuting it, or are unable to grasp this. Well, so be it.

Comment Re:I need to see more (Score 1) 711

It should be noted, though, that any spaceship travelling less than 0.1c would be surpassed by future tech that *would* go at 0.1c or faster (like a lasersail-based one can achieve). Thus, an 'interstellar arc'-ship - even with fusion-engines very slow - is not well suited for travelling to a star, and would not be recommended, unless their is some doom-like scenario threatening for the immediate extinction of the human race, and no other possibilities are left open for some reason.

Otherwise, you're just wasting enormous amounts money, time, and effort, on a risky venture to send inhabitants in such a closure for hundreds or even thousands of years, who, when arriving at their destination (a live-able planet), would notice it's already habited by other humans who departed much later, but arrived much sooner.

Comment Re:I'm no where near as smart as most of you.. (Score 1) 711

You must have missed the very, very many posts and all the long explanations as how EMdrives can break CoE as well as CoM, and thus become perpetuum mobile/free energy/over unity machines.

As a poster already told you: it's not about sending an EMdrive as described, into space with a battery-pack, but showing that an EMdrive can be made into a perpetuum mobile/free energy device, IF one presumes it's a reactionless device, as described. And the fact that the EM drive can be used to build a perpetual motion machine makes its very existence impossible. And btw, it comes over as really immature to shout 'go back and redo all your math' to people who have clearly shown they understood the basic problem better than you did.

Anyway, IF your contention is that it's possible to have a reactionless drive, yet claim that doesn't violate CoE, I have to refute that.

Let me explain:

Conservation of energy and conservation of momentum are equally fundamental. So perpetual motion machines are no more or less impossible than reactionless drives. Here's how the latter implies the former: relativity says there's no such thing as absolute velocity. There's only your velocity compared to something else. You have an infinite number of velocities at once, but can only have one acceleration. So there's no way acceleration can depend on your velocity.

So you have constant acceleration from your energy input, but your kinetic energy is going up *with the square* of your velocity, and at some point you're getting more energy out than you put in. The point at which you start getting more energy than you put in depends on the thrust to power ratio. The higher the ratio the lower the minimum speed needs to be for that to happen. A normal photon rocket has a theoretical maximum thrust to power ratio such that the speed is c (the speed of light in vacuum). It can never reach that speed, so the problem is avoided. The reported thrust to power ratios for the EmDrive are orders of magnitude higher than that, so the problem *does* exist for them.

Comment Re:I need to see more (Score 1) 711

You're not quite right in that assertion, I'm afraid. Let me explain:

IF, however, your contention is that it's possible to have a reactionless drive, yet claim that doesn't violate CoE, I have to refute that. (your position in this is not clear).

Let me explain:

Conservation of energy and conservation of momentum are equally fundamental. So perpetual motion machines are no more or less impossible than reactionless drives. Here's how the latter implies the former: relativity says there's no such thing as absolute velocity. There's only your velocity compared to something else. You have an infinite number of velocities at once, but can only have one acceleration. So there's no way acceleration can depend on your velocity.

So you have constant acceleration from your energy input, but your kinetic energy is going up *with the square* of your velocity, and at some point you're getting more energy out than you put in. The point at which you start getting more energy than you put in depends on the thrust to power ratio. The higher the ratio the lower the minimum speed needs to be for that to happen. A normal photon rocket has a theoretical maximum thrust to power ratio such that the speed is c (the speed of light in vacuum). It can never reach that speed, so the problem is avoided. The reported thrust to power ratios for the EmDrive are orders of magnitude higher than that, so the problem *does* exist for them.

Comment Re:I'm no where near as smart as most of you.. (Score 1) 711

I'll, instead, respond to your post, thank you very much. What you do with your hand is your own (and I don't have to know the yucky details ;-))

I've explained it in another post, also to you if memory serves well, very clearly and very easily to understand, why something that breaks CoM also breaks CoE, and why the EMdrive DOES break both. I can't help it if you still do not comprehend.

And while you might think it doesn't, just like you think I'm 'that other guy'; you're wrong on both accounts.

I guess I'll refrain from comparing you to a Xerox copy, not because there is no ample opportunity too, but at least I want to keep debates relevant and mildly polite (depending on how another poster is in return, of course).

Comment Re:I'm no where near as smart as most of you.. (Score 1) 711

IF, however, your contention is that it's possible to have a reactionless drive, yet claim that doesn't violate CoE, I have to refute that. (your position in this is not clear).

Let me explain:

Conservation of energy and conservation of momentum are equally fundamental. So perpetual motion machines are no more or less impossible than reactionless drives. Here's how the latter implies the former: relativity says there's no such thing as absolute velocity. There's only your velocity compared to something else. You have an infinite number of velocities at once, but can only have one acceleration. So there's no way acceleration can depend on your velocity.

So you have constant acceleration from your energy input, but your kinetic energy is going up *with the square* of your velocity, and at some point you're getting more energy out than you put in. The point at which you start getting more energy than you put in depends on the thrust to power ratio. The higher the ratio the lower the minimum speed needs to be for that to happen. A normal photon rocket has a theoretical maximum thrust to power ratio such that the speed is c (the speed of light in vacuum). It can never reach that speed, so the problem is avoided. The reported thrust to power ratios for the EmDrive are orders of magnitude higher than that, so the problem *does* exist for them.

Comment Re:I'm no where near as smart as most of you.. (Score 1) 711

Ermm... I don't know if you realize this, but if your contention is you can't derive free energy out of an EMdrive, that is exactly what I and the other guy are claiming as well...

But while you claim the problem lays in practical problems, we claim it's far more inherent then that, and is based on the principle of it.

Comment Re:I'm no where near as smart as most of you.. (Score 1) 711

No, they do not. And that's what you do not understand, I know. I've talked enough to people like you to realize this. It's like you guys have a mental block.

What the experiments showed, was the difficulty in singling out the cause of a minor force between a lot of noise, which an extremely high likelihood of being an artifact or measure error. That's really all the experiments showed.

Yet, people like you think it makes a strong case for a reactionless drive. It doesn't and isn't.

But it's impossible to make you comprehend that, because you do not *WANT* to comprehend it.

Comment Re:I'm no where near as smart as most of you.. (Score 1) 711

Wrong. Even if losses would be cumulative, the kinetic energy augments with the square of the thrust.

It's also obvious to see. If your device turns and keeps going at 1000 W, it doesn't matter *where* that 1000 W comes from. So if it comes from the device itself, it doesn't matter for the generator.

Ergo, unless you claim that the resistance also augments to the square with each given energy of the input - which is bullocks - it's obvious that at a certain point, your velocity (and thus the energy it gives) will become great enough to keep the whole thing running. At which point you have a perpetuum mobile device.

Also, as the parent poster said: debating the practicality and possible technical difficulties do not change the fact that one could actually make a perpetuum mobile. The breaking of CoM and CoE or any other basic law is not depended on technical difficulties in making a practical device that is easily built. Whether you need superconducting wires, a very low-friction axis, 100 of years of running it, high speeds of hundreds of km/h, and whether you have to pour millions of dollars in it to build it, etc. does not matter: the point is, it's possible to make one, so it breaks a basic law of physics.

Comment Re:Casimir effect (Score 1) 711

To the square?

People claiming the torsional resistance is the problem don't realize it doesn't really matter if the EM drive really is a reactionless drive. There is no increase of the resistance to the square of the input, while there IS through the velocity (kinetic energy) for every amount of input. It's not difficult to see, thus, that at some point, the energy getting out of it will be greater than whatever input + resistance will demand.

Also, people arguing about the *difficulty* of it don't seem to realize that the actual problem is, that it would actually allow a perpetuum mobile in principle; and it's that which makes it impossible. Technical difficulties are not a good argument against the principle of the matter, if that principle would actually allow a perpetuum mobile. In this case, the only ned is to augment the input enough to the point that the return, with low efficiency and all, supersedes the input and all resistances - which is possible, since the energy that one can derive from it is the SQAURE of the energy needed for the thrust. Is one really claiming resistance augments with the square too? If not, then the machine will, at a certain point, become a perpetuum mobile.

Comment Re:I need to see more (Score 1) 711

Hey, you don't have to convince me its impossible. That was just the whole point, to demonstrate the implausibility of it.

However, the impossibility of it stems from the fact that a reactionless drive, such as the EM drive, is not possible. EM fanfappers however claim it is, and I pointed out that if that were true, you would be able to make a perpetuum mobile out of it. And in that case, it wouldn't matter much how efficient the conversion is, as long as the energy getting out of it is more than the energy to run the thing. And that would be possible at some point, since the energy of velocity squares with each amount of thrust you put in (and the thrust is dependent on the energy you put in).

So torque, (in)efficiency only postpone that point, it does not inhibit it, unless one argues the torque etc. also squares with the input (which it doesn't).

So people claiming the EMdrive can't turn into a perpetuum mobile or over unity device are wrong. It can be made into one, IF the EMdrive ould be a reactionless drive. Which, of course, it isn't.

Comment Re:I need to see more (Score 1) 711

Incorrect. I do too. For the simple reason that, if the EMdrive works as it is described to work (aka, as a reactionless drive), it's the only logical conclusion to take. The only other option, if one refutes the former, is to conclude that fundamental laws vary depending on localisation. This in turn would mean, the speed of light varies, the strong nuclear force would change, etc., and thus whole swats of matter would spontaneously disintegrate into atomic and subatomic particles and exotic matter, and flood the universe . This, however, we have not observed, not even once, for the last 400 years. Hence, the extreme unlikelihood of such a claim.

You can't have it both ways. It either follows the laws of physics, in which case you can make it into an over unity device (which is extremely unlikely), or it doesn't follow the basic laws of physics, but that would entail - since we've observed those laws working to an astonishingly accurate level - that those laws vary depending on localisation and that we somehow missed the telltale signs of such a thing for the last 400 years, which is also highly unlikely.

Hence the most likely assumption, namely that the EMdrive isn't a reactionless drive at all, but the result is rather a measure error or artefact.

Comment Re:I'm no where near as smart as most of you.. (Score 1) 711

For f- sake, you simply seem NOT be able to comprehend this.

It doesn't work, because there is no such thing as a reactionless drive. If there WERE such a thing, however, like the EM drive is claimed to be, then you can get more energy out of it that you put in, by the SQUARE of it (through kinetic energy). EVEN if you have losses and the efficiency would be low, this means you can get more energy out than you put in, because those losses DO NOT rise as the square of your input.

The reason ALL PERPETUUM MOBILE and free energy devices do not work, is because you CAN NOT get more energy out of them than you put in them. Even with a frictionless and lossless conversion - which doesnt exist - it would be equal at most.

If, however, one claims one has a reactionless drive, PHYSICS say that you get the square of the energy in velocity for each amount of thrust you put in it. So IT DOES NOT MATTER whether you have losses and it isn't completely frictionless, as long as the output you have exceeds the losses. And since losses don't inversely relate in the same manner (you won't suddenly have a square loss by doubling your input), it's easy to see that, at some point, you will get more energy out than your input + losses together.

Which is, of course, not possible. Aka, the EMdrive, as described, is not possible.

For f- sake you're either dense or being wilfully obtuse.

Comment Re:I'm no where near as smart as most of you.. (Score 1) 711

You do realise that my whole point is, that such a machine can not be, right?

However, not because of losses - that would merely mean the energy you get out of it should be large enough to continue providing the required energy - but because there is no such thing as a reactionless drive to begin with, and thus the EM drive can't work as proclaimed.

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